Problem 83 - Chemistry and Chemical Reactivity

Question

Nitrogen trifluoride is prepared by the reaction of ammonia and fluorine. $$ 4 \mathrm{NH}_{3}(\mathrm{g})+3 \mathrm{F}_{2}(\mathrm{g}) \rightarrow 3 \mathrm{NH}_{4} \mathrm{F}(\mathrm{s})+\mathrm{NF}_{3}(\mathrm{g}) $$ If you mix \(\mathrm{NH}_{3}\) with \(\mathrm{F}_{2}\) in the correct stoichiometric ratio, and if the total pressure of the mixture is \(120 \mathrm{mm} \mathrm{Hg}\), what are the partial pressures of \(\mathrm{NH}_{3}\) and \(\mathrm{F}_{2} ?\) When the reactants have been completely consumed, what is the total pressure in the flask? (Assume \(T\) is constant.)

Step-by-Step Solution

Answer

Partial pressures: 68.6 mm Hg for \(\mathrm{NH}_{3}\) and 51.4 mm Hg for \(\mathrm{F}_{2}\). Final total pressure: 17.14 mm Hg.

Step 1: Identify the Stoichiometric Ratio

The balanced equation is:\[ 4 \mathrm{NH}_{3} + 3 \mathrm{F}_{2} \rightarrow 3 \mathrm{NH}_{4}\mathrm{F} + \mathrm{NF}_{3} \]This indicates that 4 moles of \(\mathrm{NH}_{3}\) react with 3 moles of \(\mathrm{F}_{2}\) to maintain a stoichiometric ratio of 4:3.

Step 2: Calculate the Partial Pressures of Reactants

The total pressure is given as 120 mm Hg. Given the stoichiometric ratio of 4:3, the partial pressure of \(\mathrm{NH}_{3}\) is \(\frac{4}{7} \times 120\) mm Hg, and for \(\mathrm{F}_{2}\) it is \(\frac{3}{7} \times 120\) mm Hg.

Step 3: Determine the Individual Partial Pressures

Calculate the partial pressures:\[ P_{\mathrm{NH}_3} = \frac{4}{7} \times 120 = 68.57 \text{ mm Hg} \approx 68.6 \text{ mm Hg}\]\[ P_{\mathrm{F}_2} = \frac{3}{7} \times 120 = 51.43 \text{ mm Hg} \approx 51.4 \text{ mm Hg}\]

Step 4: Analyze the Products and Calculate Final Pressure

Since 4 moles of \(\mathrm{NH}_{3}\) and 3 moles of \(\mathrm{F}_{2}\) react to give 1 mole of \(\mathrm{NF}_{3}\), the change in total number of moles is:\(\Delta n = (4+3) - (3+1) = 3\) moles decrease.However, since total stoichiometry decreases by this amount and maintains total gas particle consistency by reacting fully, the total pressure remains approx 17.14 mm Hg (the leftover partial pressure equivalent). The remaining gas is predominantly solid but the gaseous product gives a lower pressure accordingly compared to initial due to solid content.

Key Concepts

Partial Pressure

In a gaseous mixture, each gas contributes to the overall pressure exerted by the mixture. This is called partial pressure. Understanding partial pressure is crucial. It allows us to see how different gases behave in a closed system. Each gas in a mixture acts independently of the others in terms of pressure. It seems like it's alone, according to Dalton's Law.

Dalton's Law of Partial Pressures states that the total pressure of a gas mixture is equal to the sum of the partial pressures of individual gases. For a mixture of gases like ammonia (\( ext{NH}_3\)) and fluorine (\( ext{F}_2\)), the partial pressures aren't the same unless their amounts and molar amounts are equal.

• Formula: \( P_{ ext{total}} = P_{ ext{NH}_3} + P_{ ext{F}_2} \)
• Using stoichiometric ratios, we can find these pressures when the total pressure is known.

In our problem, the ratio of 4:3 for \( ext{NH}_3\) and \( ext{F}_2\) means that their partial pressures are derived from this ratio.

Gas Laws

Gas laws describe how gases behave under different conditions. They relate to temperature, volume, and pressure. Consider several important laws:

• Boyle's Law: Relates pressure and volume, stating that \( P_1V_1 = P_2V_2 \).
• Charles's Law: Relates volume and temperature, represented by \( V_1/T_1 = V_2/T_2 \).
• Ideal Gas Law: Combines these into one expression, \( PV = nRT \), where \( n \) is moles and \( R \) is the gas constant.

These principles help us understand changes in a reaction. In our exercise, when gases are mixed, pressure shifts result from stoichiometric reactions. Volume remains constant, thus Browne’s Law applies, where \( P \) partially depends on temperature since volume is set.

Reaction Equations

A reaction equation offers a detailed map of a chemical reaction. It shows reactants transforming into products. Writing and balancing these equations is vital in chemical stoichiometry, especially in predicting product formation.

For the given reaction:\[ 4 ext{NH}_3(g) + 3 ext{F}_2(g) \rightarrow 3 ext{NH}_4 ext{F}(s) + ext{NF}_3(g) \]

• Balancing: Keep atoms equal on both sides. The equation is already balanced, which is essential for accurate stoichiometric calculations.
• Stoichiometric Coefficients: These show the proportion of reactants and products involved, aiding in the calculation of partial pressures and amount of substance.

By adhering to balanced equations, predictions on the final state of a reaction can be accurate. This ensures chemical changes in the lab or industry are controlled and predictable.